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求函数y=log1/2[cos(x/3+π/4)]的单调递增区间.【1/2为底数】
[6kπ-(3π)/4,6kπ+(3π)/4)(k属于z)
人气:224 ℃ 时间:2019-12-15 09:04:30
解答
∵ 1 ≤ (x/3+π/4)] ≤ 1∴ 零和负数无对数∴ 0< (x/3+π/4)] ≤ 10<1/2<1∴y= log(/2) h(x)单调减即当cos(x/3+π/4)]在定义域内单调减时,y= log(/2) h(x)单调增即当2kπ ≤ x/3+π/4 ≤ 2kπ+π,即6kπ- 3π...
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